(iii) f(x−2)=(x−2)2−1=x2−4x+4−1=x2−4x+3\begin{aligned} f(x-2)&=(x-2)^2-1\\ &=x^2-4x+4-1\\ &=x^2-4x+3 \end{aligned}f(x−2)=(x−2)2−1=x2−4x+4−1=x2−4x+3 f(x−2)=x2−4x+3\boxed{f(x-2)=x^2-4x+3}f(x−2)=x2−4x+3